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Fourier Series

# Definition of Fourier Series and Typical Examples

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Problems 1-2
Page 2
Problems 3-6

Baron Jean Baptiste Joseph Fourier $$\left( 1768-1830 \right)$$ introduced the idea that any periodic function can be represented by a series of sines and cosines which are harmonically related.

To consider this idea in more detail, we should introduce some definitions and common terms.

### Basic Definitions

A function $$f\left( x \right)$$ is said to have period $$P$$ if $$f\left( {x + P} \right) = f\left( x \right)$$ for all $$x.$$ Let the function $$f\left( x \right)$$ has period $$2\pi.$$ In this case, it is enough to consider behavior of the function on the interval $$\left[ { – \pi ,\pi } \right].$$

1. Suppose that the function $$f\left( x \right)$$ with period $$2\pi$$ is absolutely integrable on $$\left[ { – \pi ,\pi } \right]$$ so that the following so-called Dirichlet integral is finite:
$\int\limits_{ – \pi }^\pi {\left| {f\left( x \right)} \right|dx} \lt \infty ;$
2. Suppose also that the function $$f\left( x \right)$$ is a single valued, piecewise continuous (must have a finite number of jump discontinuities), and piecewise monotonic (must have a finite number of maxima and minima).

If the conditions $$1$$ and $$2$$ are satisfied, the Fourier series for the function $$f\left( x \right)$$ exists and converges to the given function (see also the Convergence of Fourier Series page about convergence conditions.)

At a discontinuity $${x_0}$$, the Fouries Series converges to
$\lim\limits_{\varepsilon \to 0} \frac{1}{2}\left[ {f\left( {{x_0} – \varepsilon } \right) – f\left( {{x_0} + \varepsilon } \right)} \right].$ The Fourier series of the function $$f\left( x \right)$$ is given by
${f\left( x \right) = \frac{{{a_0}}}{2} }+{ \sum\limits_{n = 1}^\infty {\left\{ {{a_n}\cos nx + {b_n}\sin nx} \right\}} ,}$ where the Fourier coefficients $${{a_0}},$$ $${{a_n}},$$ and $${{b_n}}$$ are defined by the integrals
${{a_0} = \frac{1}{\pi }\int\limits_{ – \pi }^\pi {f\left( x \right)dx} ,\;\;\;}\kern-0.3pt {{a_n} = \frac{1}{\pi }\int\limits_{ – \pi }^\pi {f\left( x \right)\cos nx dx} ,\;\;\;}\kern-0.3pt {{b_n} = \frac{1}{\pi }\int\limits_{ – \pi }^\pi {f\left( x \right)\sin nx dx} .}$ Sometimes alternative forms of the Fourier series are used. Replacing $${{a_n}}$$ and $${{b_n}}$$ by the new variables $${{d_n}}$$ and $${{\varphi_n}}$$ or $${{d_n}}$$ and $${{\theta_n}},$$ where
${{d_n} = \sqrt {a_n^2 + b_n^2} ,\;\;\;}\kern-0.3pt {\tan {\varphi _n} = \frac{{{a_n}}}{{{b_n}}},\;\;\;}\kern-0.3pt {\tan {\theta _n} = \frac{{{b_n}}}{{{a_n}}},}$ we can write:
${f\left( x \right) = \frac{{{a_0}}}{2} }+{ \sum\limits_{n = 1}^\infty {{d_n}\sin \left( {nx + {\varphi _n}} \right)} \;\;}\kern-0.3pt{\text{or}\;\;} {f\left( x \right) = \frac{{{a_0}}}{2} }+{ \sum\limits_{n = 1}^\infty {{d_n}\cos\left( {nx + {\theta _n}} \right)} .}$

### Fourier Series of Even and Odd Functions

The Fourier series expansion of an even function $$f\left( x \right)$$ with the period of $$2\pi$$ does not involve the terms with sines and has the form:
${f\left( x \right) = \frac{{{a_0}}}{2} }+{ \sum\limits_{n = 1}^\infty {{a_n}\cos nx} ,}$ where the Fourier coefficients are given by the formulas
${{a_0} = \frac{2}{\pi }\int\limits_0^\pi {f\left( x \right)dx} ,\;\;\;}\kern-0.3pt {{a_n} = \frac{2}{\pi }\int\limits_0^\pi {f\left( x \right)\cos nxdx} .}$ Accordingly, the Fourier series expansion of an odd $$2\pi$$-periodic function $$f\left( x \right)$$ consists of sine terms only and has the form:
$f\left( x \right) = \sum\limits_{n = 1}^\infty {{b_n}\sin nx} ,$ where the coefficients $${{b_n}}$$ are
${b_n} = \frac{2}{\pi }\int\limits_0^\pi {f\left( x \right)\sin nxdx} .$ Below we consider expansions of $$2\pi$$-periodic functions into their Fourier series, supposing that these expansions exist and are convergent.

## Solved Problems

Click on problem description to see solution.

### ✓Example 1

Let the function $$f\left( x \right)$$ be $$2\pi$$-periodic and suppose that it is presented by the Fourier series:
${f\left( x \right) = \frac{{{a_0}}}{2} \text{ + }}\kern0pt{ \sum\limits_{n = 1}^\infty {\left\{ {{a_n}\cos nx + {b_n}\sin nx} \right\}}}$ Calculate the coefficients $${{a_0}},$$ $${{a_n}},$$ and $${{b_n}}.$$

### ✓Example 2

Find the Fourier series for the square $$2\pi$$-periodic wave defined on the interval $$\left[ { – \pi ,\pi } \right]:$$
${f\left( x \right) \text{ = }}\kern0pt {\begin{cases} 0, & \text{if} & – \pi \le x \le 0 \\ 1, & \text{if} & 0 < x \le \pi \end{cases}.}$

### ✓Example 3

Find the Fourier series for the sawtooth wave defined on the interval $$\left[ { – \pi ,\pi } \right]$$ and having period $$2\pi.$$

### ✓Example 4

Let $$f\left( x \right)$$ be a $$2\pi$$-periodic function such that $$f\left( x \right) = {x^2}$$ for $$x \in \left[ { – \pi ,\pi } \right].$$ Find the Fourier series for the parabolic wave.

### ✓Example 5

Find the Fourier series for the triangle wave
${f\left( x \right) \text{ = }}\kern0pt {\begin{cases} \frac{\pi }{2} + x, & \text{if} & – \pi \le x \le 0 \\ \frac{\pi }{2} – x, & \text{if} & 0 \lt x \le \pi \end{cases},}$ defined on the interval $$\left[ { – \pi ,\pi } \right].$$

### ✓Example 6

Find the Fourier series for the function
${f\left( x \right) \text{ = }}\kern0pt {\begin{cases} -1, & \text{if} & – \pi \le x \le – \frac{\pi }{2} \\ 0, & \text{if} & – \frac{\pi }{2} \lt x \le \frac{\pi }{2} \\ 1, & \text{if} & \frac{\pi }{2} \lt x \le \pi \end{cases},}$ defined on the interval $$\left[ { – \pi ,\pi } \right].$$

### Example 1.

Let the function $$f\left( x \right)$$ be $$2\pi$$-periodic and suppose that it is presented by the Fourier series:
${f\left( x \right) = \frac{{{a_0}}}{2} \text{ + }}\kern0pt{ \sum\limits_{n = 1}^\infty {\left\{ {{a_n}\cos nx + {b_n}\sin nx} \right\}}}$ Calculate the coefficients $${{a_0}},$$ $${{a_n}},$$ and $${{b_n}}.$$

#### Solution.

To define $${{a_0}},$$ we integrate the Fourier series on the interval $$\left[ { – \pi ,\pi } \right]:$$
${\int\limits_{ – \pi }^\pi {f\left( x \right)dx} } = {\pi {a_0} }+{ \sum\limits_{n = 1}^\infty {\left[ {{a_n}\int\limits_{ – \pi }^\pi {\cos nxdx} }\right.}+{\left.{ {b_n}\int\limits_{ – \pi }^\pi {\sin nxdx} } \right]}}$ For all $$n \gt 0$$,
${{\int\limits_{ – \pi }^\pi {\cos nxdx} }={ \left. {\left( {\frac{{\sin nx}}{n}} \right)} \right|_{ – \pi }^\pi }={ 0\;\;}{\text{and}\;\;\;}}\kern-0.3pt {{\int\limits_{ – \pi }^\pi {\sin nxdx} }={ \left. {\left( { – \frac{{\cos nx}}{n}} \right)} \right|_{ – \pi }^\pi }={ 0.}}$ Therefore, all the terms on the right of the summation sign are zero, so we obtain
${\int\limits_{ – \pi }^\pi {f\left( x \right)dx} = \pi {a_0}\;\;\text{or}\;\;\;}\kern-0.3pt {{a_0} = \frac{1}{\pi }\int\limits_{ – \pi }^\pi {f\left( x \right)dx} .}$ In order to find the coefficients $${{a_n}},$$ we multiply both sides of the Fourier series by $$\cos mx$$ and integrate term by term:
${\int\limits_{ – \pi }^\pi {f\left( x \right)\cos mxdx} } = {\frac{{{a_0}}}{2}\int\limits_{ – \pi }^\pi {\cos mxdx} } + {\sum\limits_{n = 1}^\infty {\left[ {{a_n}\int\limits_{ – \pi }^\pi {\cos nx\cos mxdx} }\right.}+{\left.{ {b_n}\int\limits_{ – \pi }^\pi {\sin nx\cos mxdx} } \right]} .}$ The first term on the right side is zero. Then, using the well-known trigonometric identities, we have
${\int\limits_{ – \pi }^\pi {\sin nx\cos mxdx} } = {\frac{1}{2}\int\limits_{ – \pi }^\pi {\left[ {\sin{\left( {n + m} \right)x} }\right.}+{\left.{ {\sin \left( {n – m} \right)x}} \right]dx} = 0,}$ ${\int\limits_{ – \pi }^\pi {\cos nx\cos mxdx} } = {\frac{1}{2}\int\limits_{ – \pi }^\pi {\left[ {\cos {\left( {n + m} \right)x} }\right.}+{\left.{ {\cos \left( {n – m} \right)x}} \right]dx} = 0,}$ if $$m \ne n.$$

In case when $$m = n$$, we can write:
$\require{cancel} {\int\limits_{ – \pi }^\pi {\sin nx\cos mxdx} } = {\frac{1}{2}\int\limits_{ – \pi }^\pi {\left[ {\sin 2mx + \sin 0} \right]dx} ,\;\;}\Rightarrow {\int\limits_{ – \pi }^\pi {{\sin^2}mxdx} }={ \frac{1}{2}\left[ {\left. {\left( { – \frac{{\cos 2mx}}{{2m}}} \right)} \right|_{ – \pi }^\pi } \right] } = {\frac{1}{{4m}}\left[ { – \cancel{\cos \left( {2m\pi } \right)} }\right.}+{\left.{ \cancel{\cos \left( {2m\left( { – \pi } \right)} \right)}} \right] }={ 0;}$ ${\int\limits_{ – \pi }^\pi {\cos nx\cos mxdx} } = {\frac{1}{2}\int\limits_{ – \pi }^\pi {\left[ {\cos 2mx + \cos 0} \right]dx} ,\;\;}\Rightarrow {\int\limits_{ – \pi }^\pi {{\cos^2}mxdx} } = {\frac{1}{2}\left[ {\left. {\left( {\frac{{\sin 2mx}}{{2m}}} \right)} \right|_{ – \pi }^\pi + 2\pi } \right] } = {\frac{1}{{4m}}\left[ {\sin \left( {2m\pi } \right) }\right.}-{\left.{ \sin \left( {2m\left( { – \pi } \right)} \right)} \right] + \pi }={ \pi .}$ Thus,
${\int\limits_{ – \pi }^\pi {f\left( x \right)\cos mxdx} = {a_m}\pi ,\;\;}\Rightarrow {{a_m} = \frac{1}{\pi }\int\limits_{ – \pi }^\pi {f\left( x \right)\cos mxdx} ,\;\;}\kern-0.3pt {m = 1,2,3, \ldots }$ Similarly, multiplying the Fourier series by $$\sin mx$$ and integrating term by term, we obtain the expression for $${{b_m}}:$$
${{b_m} = \frac{1}{\pi }\int\limits_{ – \pi }^\pi {f\left( x \right)\sin mxdx} ,\;\;\;}\kern-0.3pt {m = 1,2,3, \ldots }$ Rewriting the formulas for $${{a_n}},$$ $${{b_n}},$$ we can write the final expressions for the Fourier coefficients:
${{a_n} = \frac{1}{\pi }\int\limits_{ – \pi }^\pi {f\left( x \right)\cos nxdx} ,\;\;\;}\kern-0.3pt {{b_n} = \frac{1}{\pi }\int\limits_{ – \pi }^\pi {f\left( x \right)\sin nxdx} .}$

### Example 2.

Find the Fourier series for the square $$2\pi$$-periodic wave defined on the interval $$\left[ { – \pi ,\pi } \right]:$$
${f\left( x \right) \text{ = }}\kern0pt {\begin{cases} 0, & \text{if} & – \pi \le x \le 0 \\ 1, & \text{if} & 0 < x \le \pi \end{cases}.}$

#### Solution.

First we calculate the constant $${{a_0}}:$$
${{a_0} = \frac{1}{\pi }\int\limits_{ – \pi }^\pi {f\left( x \right)dx} } = {\frac{1}{\pi }\int\limits_0^\pi {1dx} } = {\frac{1}{\pi } \cdot \pi }={ 1.}$ Find now the Fourier coefficients for $$n \ne 0:$$
${{a_n} = \frac{1}{\pi }\int\limits_{ – \pi }^\pi {f\left( x \right)\cos nxdx} } = {\frac{1}{\pi }\int\limits_0^\pi {1 \cdot \cos nxdx} } = {\frac{1}{\pi }\left[ {\left. {\left( {\frac{{\sin nx}}{n}} \right)} \right|_0^\pi } \right] } = {\frac{1}{{\pi n}} \cdot 0 }={ 0,}$ ${{b_n} = \frac{1}{\pi }\int\limits_{ – \pi }^\pi {f\left( x \right)\sin nxdx} } = {\frac{1}{\pi }\int\limits_0^\pi {1 \cdot \sin nxdx} } = {\frac{1}{\pi }\left[ {\left. {\left( { – \frac{{\cos nx}}{n}} \right)} \right|_0^\pi } \right] } = { – \frac{1}{{\pi n}} \cdot \left( {\cos n\pi – \cos 0} \right) } = {\frac{{1 – \cos n\pi }}{{\pi n}}.}$ As $$\cos n\pi = {\left( { – 1} \right)^n},$$ we can write:
${b_n} = \frac{{1 – {{\left( { – 1} \right)}^n}}}{{\pi n}}.$ Thus, the Fourier series for the square wave is
${f\left( x \right) = \frac{1}{2} }+{ \sum\limits_{n = 1}^\infty {\frac{{1 – {{\left( { – 1} \right)}^n}}}{{\pi n}}\sin nx} .}$ We can easily find the first few terms of the series. By setting, for example, $$n = 5,$$ we get
${f\left( x \right) = \frac{1}{2} }+{ \frac{{1 – \left( { – 1} \right)}}{\pi }\sin x } + {\frac{{1 – {{\left( { – 1} \right)}^2}}}{{2\pi }}\sin 2x } + {\frac{{1 – {{\left( { – 1} \right)}^3}}}{{3\pi }}\sin 3x } + {\frac{{1 – {{\left( { – 1} \right)}^4}}}{{4\pi }}\sin 4x } + {\frac{{1 – {{\left( { – 1} \right)}^5}}}{{5\pi }}\sin 5x + \ldots } = {\frac{1}{2} + \frac{2}{\pi }\sin x } + {\frac{2}{{3\pi }}\sin 3x } + {\frac{2}{{5\pi }}\sin 5x + \ldots }$

Figure 1, n = 10

The graph of the function and the Fourier series expansion for $$n = 10$$ is shown in Figure $$1.$$

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Problems 1-2
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Problems 3-6